Problem: Rewrite the equation by completing the square. $2 x^{2} -9 x +7 = 0$ $(x + $
Solution: $\begin{aligned} 2 x^2 -9 x +7&=0 \\\\ 2 x^2 -9 x &=-7 \\\\ x^2 -\dfrac{9}{2} x&=-\dfrac{7}{2} \end{aligned}$ Now we want to complete $x^2 -\dfrac{9}{2} x$ into a perfect square. To do that, we should add $\left(\dfrac{{-\frac{9}{2}}}{2}\right)^2={\dfrac{81}{16}}$ to it: $x^2{-\dfrac{9}{2}}x + {\dfrac{81}{16}}=\left(x -\dfrac{9}{4} \right)^2$ $\begin{aligned} x^2 -\dfrac{9}{2} x&=-\dfrac{7}{2} \\\\ x^2 -\dfrac{9}{2} x + {\dfrac{81}{16}}&=-\dfrac{7}{2} + {\dfrac{81}{16}} \\\\ \left(x -\dfrac{9}{4} \right)^2&=\dfrac{25}{16} \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x -\dfrac{9}{4} \right)^2=\dfrac{25}{16}$